AP EAMCET · PHYSICS · Magnetic Effects of Current
A magnetic needle lying parallel to a magnetic field is turned through \(60^{\circ}\). The work done on it is \(W\). The torque required to maintain the magnetic needle in the position mentioned above is
- A \(\sqrt{3} W\)
- B \(\frac{\sqrt{3}}{2} W\)
- C \(\frac{W}{2}\)
- D \(2 W\)
Answer & Solution
Correct Answer
(A) \(\sqrt{3} W\)
Step-by-step Solution
Detailed explanation
Work done \(\begin{aligned} W & =M B\left(1-\cos 60^{\circ}\right) \\ & =\frac{M B}{2}\end{aligned}\) The torque required to maintain the magnetic needle…
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